I have the following expression


I need to find the Fourier Transform. I know for


the Shift Theorem states that the transform is $e^{ik\delta}\hat{f}(k)$, but I'm not sure how to apply the Theorem to the product above. The Fourier Transform of a product is a convolution, I know this.

My initial guess would be to apply the Shift Theorem to each term, then "bring them together" with a convolution giving

$$e^{ik\delta}\hat{f}(k)\star e^{ik\delta}\hat{g}(k)$$

That being said, it is a guess and I would like to find clarity.



If $$h(x)=f(x)g(x)\overbrace{\implies}^{FT}H(\omega)=\int_{-\infty}^\infty F(u)G(\omega-u)du$$then$$h(x+\delta)=f(x+\delta)g(x+\delta)\overbrace{\implies}^{FT}\hat{H}(\omega)=e^{j\delta \omega}\int_{-\infty}^\infty F(u)G(\omega-u)du$$

  • $\begingroup$ Just to ensure that I'm reading this correctly, it appears my guess was correct, where I would pull out the common term in my last expression? $\endgroup$ – ThatsRightJack Mar 25 at 20:47

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